Question

How do you find the exact value of `arccos(sin(pi/3))`?

Answer 1

`pi/6`

Explanation:

knowing that `sin(pi/3)=sqrt3/2`
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`arccos(sin(pi/3))=arccos((sqrt3)/2)`
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we know that `cos(pi/6)=sqrt3/2`
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so, `pi/6=arccos(sqrt3/2)`
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`arccos(sin(pi/3))=arccos((sqrt3)/2)=pi/6`

Answer 2

`arccos(sin(1/3pi))=1/6pi`

Explanation:

By definition, `cos(1/2pi-theta)=sintheta` for all `theta`

`therefore arccos (sin(1/3pi))=arccos (cos(1/2pi-1/3pi))=arccos (cos (1/6pi))=1/6pi`